Tetrad (index notation)

In Riemannian geometry, we can introduce a coordinate system over the Riemannian manifold (at least, over a chart), giving "n" coordinates

:xⁱ, i=1,...,n

for an n-dimensional manifold. Locally, at least, this gives a basis for the 1-forms, dxⁱ where d is the exterior derivative. The dual basis for the tangent space is e_i.

Now, let's choose an orthonormal basis for the fibers of T. The rest is index manipulation.

Example

Take a 3-sphere with the radius "R" and give it polar coordinates α, θ, φ.

:e(e_α)/R, :e(e_θ)/R sin(α) and :e(e_φ)/R sin(α) sin(θ)

form an orthonormal basis of T.

Call these e₁, e₂ and e₃. Given the metric η, we can ignore the covariant and contravariant distinction for T.

Then, the dreibein,

:$e\_1=R\; dalpha$:$e\_2=R\; sin\; alpha\; d\; heta$:$e\_3=R\; sin\; alpha\; sin\; heta\; dphi$.

So,

:$de\_1=0$:$de\_2=R\; cos\; alpha\; dalpha\; wedge\; d\; heta$:$de\_3=R\; (cos\; alpha\; sin\; heta\; dalpha\; wedge\; dphi\; +\; sin\; alpha\; cos\; heta\; d\; heta\; wedge\; dphi)$.

from the relation

:$d\_mathbf\{A\}\; e=de+Awedge\; e=0$,

we get

:$A\_\{12\}=-cos\; alpha\; d\; heta$:$A\_\{13\}=-cos\; alpha\; sin\; heta\; dphi$:$A\_\{23\}=-cos\; heta\; dphi$.

(d_Aη=0 tells us A is antisymmetric)

So, $mathbf\{F\}=dmathbf\{A\}+mathbf\{A\}wedge\; mathbf\{A\}$,

:$F\_\{12\}=sinalpha\; dalphawedge\; d\; heta$:$F\_\{13\}=sin\; alpha\; sin\; heta\; dalphawedge\; dphi$:$F\_\{23\}=sin^2\; alpha\; sin\; heta\; d\; hetawedge\; dphi$